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27 August, 05:03

A crate of mass 190 kg sits on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, and the coefficient of kinetic friction is 0.36. You push the crate with a horizontal force of 500 N. What is the magnitude of the frictional force on the crate by the floor in this situation?[Use g = 9.81 m/s2]A small asteroid with a mass of 2000 kg moves near the Earth. At a particular instant the asteroid's velocity is 〈 - 1.30 x 104, 4.20 x 104, 0 〉 m/s, and its position with respect to the center of the center of the Earth is 〈 6.00 x 106, 10.00 x 106, 0 〉 m. The center of the Earth (mass = 5.97 x 1024 kg), is at the origin of the coordinate system. What is the (approximate) new momentum of the asteroid 1.50 x 103 seconds later?[Use: G = 6.67 x 10-11 Nm2 / kg2 ]

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  1. 27 August, 05:20
    Mass of 190kg

    Coefficient of static friction is 0.4

    Coefficient of kinetic friction 0.36

    Horizontal force = 500N

    Taking g=9.81m/s^2.

    The weight of the body my


    There is a normal acting on the body which is equal to the weight


    Frictional force (fr) is acting on the body and it is opposite the horizontal force.

    The minimum force to be overcome before the object can start to move is Fr = μsN

    Fr = μsN. μs=0.4

    Fr = 0.4*1863.91


    Since the horizontal force (500N) is not up to the minimum force to make the object move, then the force of 500N the body is still at rest.

    Then the frictional force at that time is equal to the horizontal force


    Functional force = 500N

    b. Mass of asteroid is


    Asteroid velocity at a particular instant is,

    U = (-1.30x10^4, 4.20x10^4, 0) m/s

    Magnitude of U is

    U=√ (-1.30*10^4) ^2 + (4.2*10^4) ^2+0



    Position of the asteroid from the centre of the earth is,

    R = (6.00x10^6, 10.00x10^6, 0) m.

    The magnitude of the radius is

    R = √ (6.00x10^6) ^2 + (10.00x10^6) ^2 + 0^2





    The mass of the earth is

    Me=5.97x10^24 kg

    The momentum of the asteroid after time, t=1.5*10^3s

    Given that G=6.67x10^-11Nm^2/kg^2

    Momentum is


    There the new momentum will be


    Now we the to find the force the earth exert on the asteroid by using


    F=6.67E-11 * 2000 * 5.97E24 / 13.6E13



    The new momentum

    Mv = Mu+Ft

    Mv = 2000 (4.39E4) + 5855.88 (1.5E3)


    The new momentum is 9.66*10^7 Kgm/s
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